To find the Highest Common Factor (HCF) of 52 and 117 and express it in the form 52x + 117y, we can use the Euclidean algorithm.

1. Step 1: Find the Remainder

   Divide the larger number by the smaller number and find the remainder.

   117=2×52+13117=2×52+13

   So, the remainder is 13.

2. Step 2: Replace Numbers

   Now, replace the divisor with the previous remainder, and the dividend with the divisor.

   52=4×13+052=4×13+0

   Here, the remainder is 0, so we stop. The divisor at this step, which is 13, is the HCF.

3. Expressing in the given form

   Now, we backtrack to express the HCF, which is 13, in the form 52x + 117y.

   Using the reverse steps of the Euclidean algorithm, we can express the HCF as a linear combination of 52 and 117:

   13=117−2×5213=117−2×52

   Hence, the HCF of 52 and 117 expressed in the form 52x + 117y is 13=117−2×5213=117−2×52.

To prove that x2−xx2−x is divisible by 2 for all positive integer xx, we can use mathematical induction.

Base Case: Let's start by checking the base case. When x=1x=1, x2−x=12−1=0x2−x=12−1=0. Since 0 is divisible by 2, the base case holds.

Inductive Hypothesis: Assume that x2−xx2−x is divisible by 2 for some positive integer kk, i.e., k2−kk2−k is divisible by 2.

Inductive Step: We need to show that if the statement holds for kk, then it also holds for k+1k+1.

(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k

By the inductive hypothesis, we know that k2−kk2−k is divisible by 2. And we know that kk is a positive integer, so kk is also divisible by 2 or it is an odd number.

• If kk is divisible by 2, then k2−kk2−k is divisible by 2, and kk is divisible by 2, so their sum (k2−k)+k(k2−k)+k is also divisible by 2.

• If kk is an odd number, then k2−kk2−k is still divisible by 2, and when you add kk to it, you still get an even number, which is divisible by 2.

In both cases, (k+1)2−(k+1)(k+1)2−(k+1) is divisible by 2.

Therefore, by mathematical induction, we conclude that x2−xx2−x is divisible by 2 for all positive integers xx.

Sure, let's prove this by induction.

Base Case:
When n=2n=2, a2=a×aa2=a×a, where aa is a positive rational number. The product of two rational numbers is rational, so a2a2 is rational.

Inductive Step:
Assume that anan is rational for some positive integer n>1n>1.

Consider an+1an+1: an+1=an×aan+1=an×a

By the inductive hypothesis, anan is rational. And since aa is rational (given in the problem), the product of two rational numbers is rational. Therefore, an+1an+1 is rational.

By mathematical induction, anan is rational for all positive integers n>1n>1.

are irrational numbers.